The circle S equiv x^2+y^2-2x-4y+1=0 cuts the | vedclass

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(A) Given,$S \equiv x^2+y^2-2x-4y+1=0$. For the $y$-axis,set $x=0$,which gives $y^2-4y+1=0$. Solving for $y$,we get $y = \frac{4 \pm \sqrt{16-4}}{2} =

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$7+2\sqrt{3} : -7+2\sqrt{3}$

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(A) Given,$S \equiv x^2+y^2-2x-4y+1=0$. For the $y$-axis,set $x=0$,which gives $y^2-4y+1=0$. Solving for $y$,we get $y = \frac{4 \pm \sqrt{16-4}}{2} = 2 \pm \sqrt{3}$. Since $OA > OB$,we have $A(0, 2+\sqrt{3})$ and $B(0, 2-\sqrt{3})$. The radical axis of $S=0$ and $S^{\prime}=0$ is given by $S-S^{\prime}=0$. $(x^2+y^2-2x-4y+1) - (x^2+y^2-4x-2y+4) = 0 \Rightarrow 2x-2y-3=0$. For the $y$-axis,set $x=0$,which gives $-2y-3=0 \Rightarrow y = -\frac{3}{2}$. So,$C$ is $(0, -\frac{3}{2})$. Let the ratio in which $C$ divides $AB$ be $k:1$. Using the section formula for the $y$-coordinate: $-\frac{3}{2} = \frac{k(2-\sqrt{3}) + 1(2+\sqrt{3})}{k+1}$. $-3(k+1) = 2(k(2-\sqrt{3}) + 2+\sqrt{3})$. $-3k-3 = 4k - 2k\sqrt{3} + 4 + 2\sqrt{3}$. $-3k-4k + 2k\sqrt{3} = 4+3+2\sqrt{3}$. $k(2\sqrt{3}-7) = 7+2\sqrt{3}$. $k = \frac{7+2\sqrt{3}}{2\sqrt{3}-7} = \frac{7+2\sqrt{3}}{-(7-2\sqrt{3})}$. Thus,the ratio $k:1$ is $(7+2\sqrt{3}) : (-7+2\sqrt{3})$.

The circle $S \equiv x^2+y^2-2x-4y+1=0$ cuts the $y$-axis at $A, B$ $(OA > OB)$. If the radical axis of $S=0$ and $S^{\prime} \equiv x^2+y^2-4x-2y+4=0$ cuts the $y$-axis at $C$,then the ratio in which $C$ divides $AB$ is:

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